Quasiconvex bulk and surface energies with subquadratic growth

1.   Introduction

Consider the following $m$th degree homogeneous polynomial of $n$ variables $f(x)$ as

where $x = (x_{1}, x_{2}, \cdots, x_{n})\in R^n$. When $m$ is even, $f(x)$ is called positive definite if

The homogeneous polynomial $f(x)$ in $(1.1)$ can be expressed as the tensor product of a symmetric tensor $\mathcal{A}$ with $m$-order, $n$-dimension and $x^{m}$ defined by

where

$C(R)$ presents complex (real) number fields. The symmetric tensor $\mathcal {A}$ is called positive definite if $f(x)$ in $(1.2)$ is positive definite ^[1]. Moreover, a tensor $\mathcal {I} = (\delta_{i_{1}i_2\cdots i_{m}})$ is called the unit tensor^[2], where

The positive definiteness of tensor has received much attention of researchers' in recent decade ^[3,4,5]. Based on the Sturm theorem, the positive definiteness of a multivariate polynomial form can be checked for $n\leq3$ ^[6]. For $n > 3$ and $m\geq 4$, it is difficult to determine the positive definiteness of $f(x)$ in $(2)$. Ni et al.^[1] provided an eigenvalue method for identifying positive definiteness of a multivariate form. However, all the eigenvalues of the tensor are needed in this method, thus the method is not practical when tensor order or dimension is large.

Recently, based on the criteria of $\mathcal {H}$-tensors, Li et al.^[7] provided a practical method for identifying the positive definiteness of an even-order symmetric tensor. $\mathcal {H}$-tensor is a special kind of tensors and an even order symmetric $\mathcal {H}$-tensor with positive diagonal entries is positive definite. Due to this, we may identify the positive definiteness of a tensor via identifying $\mathcal {H}$-tensor. For the latter, with the help of generalized diagonally dominant tensor, various criteria for $\mathcal {H}$-tensors and $\mathcal {M}$-tensors is established ^{[8,9,10,11,12,13,14,15,16]}, which only depends on the elements of the tensors and is more effective to determine whether a given tensor is an $\mathcal {H}$-tensor ($\mathcal {M}$-tensor) or not. For example, the following result is given in ^[16]:

Theorem 1. Let $\mathcal {A} = (a_{i_1\cdots i_m})$ be a complex tensor with $m$-order, $n$-dimension. If

then $\mathcal{A}$ is an $\mathcal{H}$-tensor.

In this paper, we continue to present new criterions based on $\mathcal{H}$-tensors for identifying positive definiteness of homogeneous polynomial forms. The obtained results extend the corresponding conclusions ^[16,17,18]. The validity of our proposed methods are theoretically guaranteed and the numerical experiments show their effciency.

2.   Preliminaries

In this section, some notation, definitions and lemmas are given.

Let $S$ be a nonempty subset of $N = \{1, 2, \cdots, n\}$ and let $N \backslash S$ be the complement of $S$ in $N$. Given an $m$-order $n$-dimension complex tensor $\mathcal {A} = (a_{i_1\cdots i_m})$, we denote

In this paper, we always assume that neither $N_1$ or $N_2$ is empty. Otherwise, we assume that $\mathcal {A}$ satisfies: $a_{ii\cdots i}\neq 0, \Lambda_i(\mathcal {A})\neq 0, \forall i\in N.$

we may define the following structured tensors extended from matrices.

Definition 1. ^[10] Let $\mathcal {A} = (a_{i_1i_2\cdots i_m})$ be an $m$-order $n$-dimension complex tensor. $\mathcal{A}$ is called an $\mathcal{H}$-tensor if there is a positive vector $x = (x_1, x_2, \cdots, x_n)^{T}\in R^n$ such that

Definition 2. ^[2] An $m$-order $n$-dimension complex tensor $\mathcal {A} = (a_{i_1i_2\cdots i_m})$ is called reducible if there exists a nonempty proper index subset $I\subset N$ such that

Otherwise, we say $\mathcal{A}$ is irreducible.

Example 1. Consider the $4$-order $4$-dimension tensor $\mathcal{A}$ given

and zero elsewhere. Then $a_{i_{1}i_{2}i_{3} i_{4}} = 0$ for all $i_1\in \{1, 4\}$ and for all $i_2, i_3, i_4\in \{2, 3\}$. From Definition $2$, we have that $\mathcal {A}$ is reducible.

Definition 3. ^[12] Let $\mathcal {A} = (a_{i_1i_2\cdots i_m})$ be an $m$-order $n$-dimension complex tensor, for $i, j\in N (i\neq j)$, if there exist indices $k_{1}, k_{2}, \cdots, k_{r}$ with

where $k_{0} = i, k_{r+1} = j$, we call that there is a nonzero elements chain from $i$ to $j$.

It is shown that for any $\mathcal {H}$-tensor, there exists at least one strictly diagonally dominant row ^[7]. Further, we have the following conclusion.

Lemma 1. ^[10] If $\mathcal{A}$ is a strictly diagonally dominant tensor, then $\mathcal {A}$ is an $\mathcal {H}$-tensor.

Lemma 2. ^[7] Let $\mathcal {A} = (a_{i_1\cdots i_m})$ be a complex tensor with $m$-order, $n$-dimension. If there exists a positive diagonal matrix $X$ such that $\mathcal {A}X^{m-1}$ is an $\mathcal {H}$-tensor, then $\mathcal{A}$ is an $\mathcal {H}$-tensor.

Lemma 3. ^[7] Let $\mathcal {A} = (a_{i_1\cdots i_m})$ be a complex tensor with $m$-order, $n$-dimension. If $\mathcal {A}$ is irreducible,

and strictly inequality holds for at least one $i$, then $\mathcal{A}$ is an $\mathcal {H}$-tensor.

Lemma 4. ^[12] Let $\mathcal{A} = (a_{i_{1}\cdots i_{m}})$ be a complex tensor with $m$-order, $n$-dimension. If

● $(i)$ $|a_{ii\cdots i}|\geq \Lambda_i(\mathcal{A}), \; \; \; \; \forall i\in N$,

● $(ii)$ $N_3 = \{i\in N: |a_{ii\cdots i}| > \Lambda_i(\mathcal{A}) \}\neq \emptyset$,

● $(iii)$ For any $i\notin N_3$, there exists a nonzero elements chain from $i$ to $j$ such that $j\in N_3$,

then $\mathcal {A}$ is an $\mathcal {H}$-tensor.

3.   Criteria for identifying $\mathcal {H}$-tensors

In this section, we give some new criteria for $\mathcal{H}$-tensors.

Theorem 2. Let $\mathcal{A} = (a_{i_{1}\cdots i_{m}})$ be a complex tensor with $m$-order, $n$-dimension. If for $i\in N_2$,

and for $i\in N_1$, $|a_{ii\cdots i}|\neq \sum\limits_{i_2i_3\cdots i_m\in N_0^{m-1} \atop\delta_{ii_2\cdots i_m} = 0} |a_{ii_2\cdots i_m}|$, then $\mathcal {A}$ is an $\mathcal{H}$-tensor.

Proof. From the definition of $q$, we know that $0\leq q < 1$, $q \geq \frac{\Lambda_i(\mathcal{A})-|a_{ii\cdots i}|}{\Lambda_i(\mathcal{A}) }(\forall i \in N_2)$, so for any $i\in N_3$,

that is

By the definition of $P_i(\mathcal{A})$, we have

For any $i\in N_{3}$, from Inequality (3.2) and $0\leq t \leq 1$, we conclude that

For any $i \in N _2$, by Inequality (3.1), it holds that

By Inequality (3.3) and Inequality (3.4), there exists a sufficiently small positive number $\varepsilon$ such that

and

that is,

By the definition of $t$, it holds that

that is

Let the matrix $D = diag(d_1, d_2, \cdots, d_n)$, and denote $\mathcal {B} = \mathcal {A}D^{m-1} = (b_{i_1i_2\cdots i_m}),$ where

For any $i\in N_1$, by $q > \frac{tP_{i}(\mathcal {A}) }{|a_{ii\cdots i}|}(\forall i\in N_{3})$, we conclude that

For $\forall i\in N_2$, by Inequality (3.6), then

Finally, for any $i\in N_3$, by Inequality (3.7), thus

Therefore, we obtain that $|b_{ii\cdots i}| > \Lambda_i(\mathcal {B}) (\forall i\in N)$. From Lemma 1, $\mathcal {B}$ is an $\mathcal {H}$-tensor. Further, by Lemma 2, $\mathcal {A}$ is an $\mathcal {H}$-tensor.

Remark 1. From Theorem $2$, we conclude that $0\leq p < 1$, $0\leq t\leq1$, and for any $i\in N_3$,

Thus, all conditions in Theorem $2$ are weaker than that in Theorem $1$. Example $2$ illustrates the superiority of Theorem $2$.

Theorem 3. Let $\mathcal{A} = (a_{i_{1}\cdots i_{m}})$ be a complex tensor with $m$-order, $n$-dimension. If $\mathcal {A}$ is irreducible, and for all $i\in N_2,$

and at least one strict inequality in $(3.8)$ holds, then $\mathcal {A}$ is an $\mathcal{H}$-tensor.

Proof. Notice that $\mathcal {A}$ is irreducible, this implies that for any $i \in N_3$, $P_{i}(\mathcal {A}) > 0\text{, }\; t > 0$ (Otherwise, $\mathcal{A}$ is reducible).

For any $i \in N_2$, by Inequality (3.8), we obtain

Let the matrix $D = diag(d_1, d_2, \cdots, d_n)$, denote $\mathcal {B} = \mathcal {A}D^{m-1} = (b_{i_1i_2\cdots i_m}),$ where

For any $i\in N_1$, by $q > \frac{tP_{i}(\mathcal {A})}{|a_{ii\cdots i}|}(\forall i\in N_{3})$, we conclude that

For any $i\in N_2$, by Inequality $(3.9)$, it holds that

Next, for any $i\in N_3$, by Inequality $(3.7)$, then

Therefore, $|b_{ii\cdots i}|\geq \Lambda_i(\mathcal {B})$ ($\forall i\in N$), and for all $\forall i\in N_2$, at least one strict inequality in (10) holds, that is, there exists an $i_0\in N_2$ such that $|b_{i_0i_0\cdots i_0}| > \Lambda_{i_0}(\mathcal {B})$.

On the other hand, since $\mathcal {A}$ is irreducible and so is $\mathcal {B}$. Then, by Lemma 3, we have that $\mathcal {B}$ is an $\mathcal {H}$-tensor. By Lemma 2, $\mathcal {A}$ is also an $\mathcal {H}$-tensor.

Let

Theorem 4. Let $\mathcal{A} = (a_{i_{1}\cdots i_{m}})$ be a complex tensor with $m$-order, $n$-dimension. For any $i\in N_2$,

and if for any $i \in N\setminus K(\mathcal{A})\neq \emptyset$, there exists a nonzero elements chain from $i$ to $j$ such that $j\in K(\mathcal{A})\neq \emptyset$, then $\mathcal {A}$ is an $\mathcal {H}$-tensor.

Proof. Let the matrix $D = diag(d_1, d_2, \cdots, d_n)$, and denote $\mathcal {B} = \mathcal {A}D^{m-1} = (b_{i_1i_2\cdots i_m}),$ where

A similar argument to that of Theorem 2, we can prove that $|b_{ii\cdots i}|\geq \Lambda_i(\mathcal {B}) (\forall i\in N)$, and there exists at least an $i \in N_2$ such that $|b_{ii\cdots i}| > \Lambda_i(\mathcal {B})$.

On the other hand, if $|b_{ii\cdots i}| = \Lambda_i(\mathcal {B})$, then $i \in N\setminus K(\mathcal{A})$, by the assumption, we know that there exists a nonzero elements chain of $\mathcal {A}$ from $i$ to $j$, such that $j\in K(\mathcal{A})$. Hence, there exists a nonzero elements chain of $\mathcal {B}$ from $i$ to $j$, such that $j$ satisfying $|b_{jj\cdots j}| > \Lambda_j(\mathcal {B})$.

Based on above analysis, we get that $\mathcal {B}$ satisfies the conditions of Lemma 4, so $\mathcal {B}$ is an $\mathcal {H}$-tensor. By Lemma 2, $\mathcal {A}$ is an $\mathcal {H}$-tensor.

Example 2. Consider the $3$-order $3$-dimension tensor $\mathcal {A} = (a_{ijk})$ defined as follows:

Obviously,

so $N_1 = \emptyset, N_2 = \{1\}, N_3 = \{2, 3\}$. By calculations, we have

When $i = 1$, we get

so $\mathcal {A}$ satisfies the conditions of Theorem $2$, then $\mathcal {A}$ is an $\mathcal {H}$-tensor. However,

so $\mathcal {A}$ does not satisfy the conditions of Theorem $1$.

4.   An application: the positive definiteness of homogeneous polynomial forms

Based on the criteria of $\mathcal {H}$-tensors in Section $3$, we present some criteria for identifying the positive definiteness of an even-order real symmetric tensor. First, we recall the following lemma.

Lemma 5. ^[7] Let $\mathcal{A} = (a_{i_{1}i_2\cdots i_{m}})$ be an even-order real symmetric tensor with $m$-order, $n$-dimension, and $a_{k\cdots k} > 0$ for all $k \in N$. If $\mathcal {A}$ is an $\mathcal{H}$-tensor, then $\mathcal {A}$ is positive definite.

From Theorems $2-4$ and Lemma $5$, we obtain easily the following result.

Theorem 5. Let $\mathcal{A} = (a_{i_{1}i_2\cdots i_{m}})$ be an even-order real symmetric tensor with $m$-order, $n$-dimension, and $a_{ii\cdots i} > 0$ for all $i \in N$. If one of the following holds:

● $(i)$ $\mathcal {A}$ satisfies all the conditions of Theorem $2$,

● $(ii)$ $\mathcal {A}$ satisfies all the conditions of Theorem $3$,

● $(iii)$ $\mathcal {A}$ satisfies all the conditions of Theorem $4$,

then $\mathcal {A}$ is positive definite.

Example 3. Let

be a $4$th-degree homogeneous polynomial. We can get the $4$-order $4$-dimension real symmetric tensor $\mathcal{A} = (a_{i_{1}i_{2}i_{3} i_{4}})$, where

and zero elsewhere. By calculations, we have

and

Then $\mathcal{A}$ is not strictly diagonally dominate as defined in ^[17] or quasidoubly strictly diagonally dominant as defined in ^[18]. Hence, we cannot use Theorem 3 in ^[17] and Theorem 4 in ^[18] to identify the positive definiteness of $\mathcal{A}$. However, it can be verified that $\mathcal {A}$ satisfies all the conditions of Theorem $2$.

so $N_1 = \emptyset, N_2 = \{1\}, N_3 = \{ 2, 3, 4\}$. By calculations, we have

When $i = 1$, we get

Therefore, from Theorem $5$, we have that $\mathcal {A}$ is positive definite, that is, $f(x)$ is positive definite.

5.   Conclusions

In this paper, we given some inequalities to identify whether a tensor is an $\mathcal {H}$-tensor, which was also used to identify the positive definiteness of an even degree homogeneous polynomial $f(x)\equiv\mathcal{A}x^{m}$. These inequalities were expressed in terms of the elements of $\mathcal{A}$, so they can be checked easily.

Acknowledgments

The authors wish to give their sincere thanks to the anonymous referees for their valuable suggestions and helpful comments, which help improve the quality of the paper significantly. This work was supported by the National Natural Science Foundation of China (11861077), the Foundation of Science and Technology Department of Guizhou Province (20191161, 20181079), the Talent Growth Project Department of Guizhou Province ([2016]168) and the Research Foundation of Guizhou Minzu University (2019YB08).

Conflict of interest

The authors declare that they have no competing interests.